Methods for removing air pollutants from combustion flue gas

ABSTRACT

A method is provided for preventing the release of air pollutants with combustion flue gases emitted to the atmosphere by the removal of one or more of nitric oxide, sulfur trioxide, light hydrocarbons, carbon monoxide, and trace amounts of mercury from combustion flue gas streams. The method converts nitric oxide to nitrogen dioxide, sulfur trioxide to sulfur dioxide, removes light hydrocarbons in the form of carbon dioxide, reduces the concentration of carbon monoxide, and removes mercury vapor in the form of mercury oxide, by the addition of hydrogen peroxide or a mixture of hydrogen peroxide and methanol to a combustion flue gas at a temperature in the range from about 650 K (377° C.) to 1100 K (827° C.).

This application is a continuation of U.S. application Ser. No.08/311,353, filed Sep. 23, 1994, for Methods for Removing Air Pollutantsfrom Combustion Flue Gas, abandoned.

BACKGROUND

1. Field of the Invention

The present invention relates to methods for removal of air pollutantssuch as nitric oxide (NO), sulfur trioxide (SO₃), light hydrocarbons (C₁-C₄), carbon monoxide (CO), and trace amounts of mercury (Hg) fromcombustion flue gas. More particularly, the invention relates to methodsfor conversion of nitric oxide to nitrogen dioxide (NO₂), sulfurtrioxide to sulfur dioxide (SO₂), removal of hydrocarbons in the form ofcarbon dioxide (CO₂), reduction of carbon monoxide, and removal ofmercury vapor in the form of mercury oxide (HgO) by the addition ofhydrogen peroxide or a mixture of hydrogen peroxide and methanol tocombustion flue gas under certain conditions.

2. The Relevant Technology

The major air pollutants emanating from boilers, furnaces, incinerators,engines and other combustion sources, are NO_(x), SO_(x), carbonmonoxide (CO), and various carbon containing compounds. NO_(x) isproduced in the form of nitric oxide (NO). Some nitrogen dioxide (NO₂)is also formed, but its concentration is less than 5% of the totalNO_(x) which is typically 200-1000 ppm. The combustion of coal, oil andother sulfur containing fuels produces a flue gas in which 98-99% of thesulfur is in the form of sulfur dioxide (SO₂) and 1-2% is sulfurtrioxide (SO₃). For low and high sulfur coals the total concentration ofSO_(x) is usually in the range of 1,000-4,000 ppm. The concentrations ofCO and unburned hydrocarbons depend on combustion regimes and can reachseveral hundred ppm and more.

The above listed air pollutants are the subject of growing concernbecause they are either toxic compounds or the precursors to acid raindeposition and photochemical smog. A variety of technologies have beendeveloped to reduce the pollutant emissions, but these technologies havesubstantial disadvantages and a need exists to develop novel conceptsfor improving the effectiveness of air pollution control and forreducing its cost. Some problems with the existing post-combustion airpollution control technologies are considered below.

If it is necessary to reduce the emissions of NO_(x) from a boiler orfurnace below what can be achieved by combustion modifications, thereare two kinds of post-combustion NO_(x) control technologies: SelectiveCatalytic Reduction (SCR) and Selective Non-Catalytic Reduction (SNCR).SCR processes are based on the reaction of NO and NO₂ with ammonia inthe presence of a catalyst forming nitrogen and water. These methods areeffective within a narrow flue gas temperature window which depends onthe catalyst. Other concerns regarding the application of SCR are thefollowing: high cost, risk of catalyst poisoning and thermal shock,spent catalyst disposal, and risk of ammonia leakage to the atmosphere.SNCR methods include the injection of NH_(i) radical precursors: ammonia(Thermal DeNO_(x) process), urea or cyanuric acid at temperatures ofabout 1250 K. SNCR also involves the reactions of NH₃ or its derivativeswith NO_(x), but in the absence of catalysts. The main areas of concernare ammonia breakthrough, a narrow operating temperature window, andreduced efficiency of NO_(x) removal in the presence of CO.

There are other non-catalytic post-combustion NO_(x) control methodsthat are currently under development, such as conversion of NO into NO₂followed by simultaneous scrubbing of NO₂ and SO₂ from combustion fluegas. Since flue gas desulfurization (FGD) systems are required for SO₂removal after combustion of sulfur containing fuels and it has beenproven that NO₂ can be removed efficiently in sodium-based wet scrubbersor in modified calcium-based SO₂ scrubbers, the conversion of NO intoNO₂ becomes a promising strategy for combined NO_(x) and SO_(x) removal.

Wet scrubbing methods of NO_(x) reduction are limited by the relativelyinert nature of NO--. This difficulty can be overcome by oxidation of NOto much more reactive NO₂. Oxygen is not useful for NO oxidation at fluegas conditions because their interaction is very slow and characteristictimes of their reaction are several orders of magnitude larger than theflue gas residence times in post-combustion channels. An attractivemethod of NO to NO₂ oxidation by injection of methanol (CH₃ OH) intoflue gas was invented by R. K. Lyon, Method for Preventing Formation ofSulfuric Acid and Related Products in Combustion Effluents, U.S. Pat.No. 4,849,192, which is incorporated herein by reference (hereinafterreferred to as the "Lyon Patent"). Methanol reacts with NO by means of achain reaction, and the main elementary step for NO conversion is therapid reaction of NO with HO₂ radicals:

    NO+HO.sub.2 →NO.sub.2 +OH                           (1)

Unfortunately, a problem with using methanol is the formation of CO as aby-product. Each molecule of NO converted into NO₂ produces a moleculeof CO, and at relatively low methanol injection temperatures, CO is notoxidized to CO₂.

Azuhata et al., Effect of H₂ O₂ on Homogeneous Gas Phase NO ReductionReaction with NH₃, AIChE J. v. 28, No. 1, p. 7 (1982), which isincorporated herein by reference, presented data on NO-to-NO₂ conversionby interaction with hydrogen peroxide, but the reaction times (t_(r)) intheir experiments were approximately 12 seconds, which is not applicableto air pollution control.

There are a variety of effective dry and wet FGD technologies based onscrubbing by or injection of basic compounds, such as lime, limestone,sodium salts, etc. From an air pollution control viewpoint the bestexisting FGD technologies are quite satisfactory although a need of acheaper process always exists. But in terms of boiler/furnace operation,the production of small amounts of SO₃ is a problem. When the flue gascools, the SO₃ reacts with water to form a mist of sulfuric acid whichcauses corrosion of equipment. Another SO₃ problem is that if theThermal DeNO_(x) process is used for NO_(x) removal from SO_(x)-containing flue gas, the remaining NH₃ reacts with SO₃ and water toform NH₄ HSO₄, a sticky and highly corrosive compound that can alsocause plugging. Methanol injection can successfully handle the SO₃problems because CH₃ OH does a double job: it simultaneously converts NOto NO₂ and SO₃ to SO₂. The principal elementary reaction for SO₃ -to-SO₂conversion is the interaction of SO₃ with HO₂ radicals followed by HSO₃decomposition:

    SO.sub.3 +HO.sub.2 →HSO.sub.3 +O.sub.2              ( 2)

    HSO.sub.3 +M→SO.sub.2 +OH+M                         (3)

Just as for NO to NO₂ conversion, HO₂ radicals are the active speciesfor SO₃ removal. The only remaining problem in the use of methanol is COproduction. Therefore, it would be of great benefit to find anotheradditive instead of methanol to convert SO₃ to SO₂ without formingdangerous by-products.

There is a relationship between NO and CO emissions that plays animportant role in air pollution control: the higher the peak combustiontemperature, the higher the concentration of NO that is formed, thelower CO production. At relatively low combustion temperatures, NOproduction is minimized but increased levels of CO form. At high enoughtemperature, in ideal mixing conditions and in an excess of air, all COwould be oxidized to CO₂, but the combustion processes in variousapplications are far from ideal conditions and the problem of COemission still exists. The main elementary reaction that converts CO toCO₂ in combustion gases is

    CO+OH→CO.sub.2 +H                                   (4)

Therefore, additional concentrations of hydroxyl radicals can improve COremoval, and chemical additives producing OH radicals can be useful.

According to chemical kinetics calculations, CO and hydrocarbons burnout in the excess of air, but in practical applications they are formedor not burned out. Emissions of hydrocarbons are dangerous for humanhealth and reduce combustion efficiency. As does CO, hydrocarbons reactrapidly with OH radicals in the gas phase. These reactions producedifferent organic radicals which are oxidized by O₂. Thus, OH radicalscan enhance the removal of organic compounds.

Mercury emissions from combustion sources have recently been a growingenvironmental concern. High mercury concentrations have been found invarious locations and objects. For example, fish tested from a largenumber of lakes have been found to contain mercury at levels that makethe fish unsuitable to eat. As these lakes do not have a local source ofmercury, the cause of the high levels of mercury are most likely fromdirect deposition from the air.

Currently, elemental mercury (Hg) and different forms of gaseous andparticulate mercury compounds, such as mercury oxide (HgO), mercurychlorides (HgCl₂ and Hg₂ Cl₂), etc., are emitted to the atmosphere fromvarious plants having combustion sources. If those plants are equippedwith a particulate control device, such as dry or wet electrostaticprecipitators (ESP), most of the oxidized mercury compounds are removedas particles. Elemental mercury, however, is the most volatile mercuryform, and increases the atmospheric background concentration whenemitted. It has been reported that mercury emissions fromwaste-to-energy plants range from 0 to about 100 ppb. The averageconcentration of mercury in flue gases from coal combustion plants inthe United States is expected to be about 3 ppb.

Presently, there are two groups of methods for mercury vapor removal.The first group of methods relates to adsorption or absorption by solidor liquid substances. One approach in this first group has been to treatthe entire flow of flue gases by passage through adsorbents orabsorbents. Another approach in this group of methods has been to injectcarbon into a flue gas stream in order to effect mercury adsorption. Thesecond group of methods relates to chemical conversions to non-volatilemercury compounds followed by their collection in a particulate controldevice. For example, there are chemical methods of mercury removal whichinvolve the injection of a sodium polysulfide solution into flue gas andpassing the flue gas through a so-called "selenium filter". Thesemethods were recently described by Dismukes, E. B., Trace ElementControl in Electrostatic Precipitators and Fabric Filters, Proc. of theInternational Workshop "Trace Elements Transformations in Coal-FiredPower Systems", Scottsdale, Ariz., April 1993, which is incorporatedherein by reference. All existing methods of mercury removal areexpensive and not highly effective.

At combustion temperatures, all mercury is probably in the form of atomsbecause the mercury compounds are not very stable. As the flue gasescool and reach temperatures between 600° and 700° C., the oxidation ofmercury to HgO can occur. It has been found, however, that oxygen itselfis not a suitable oxidant under flue gas conditions. The reaction ofmercury with oxygen that is usually present in flue gas, Hg+0.5O₂ =HgO,is too slow to remove a significant amount of mercury.

As was described by Masuda, et al., Pulse Corona Induced Plasma ChemicalProcess for DeNO_(x), DeSO_(x) and Mercury Vapor Control of CombustionGas, Proc. of the 3d Intern. Conf. on Electrostatic Precipitation,Abano-Padova, Italy, October 1987, pp. 667-676, which is incorporatedherein by reference, in the presence of corona discharge, an efficientmercury removal of up to 90+% can be achieved. In this case, mercuryreacts with more energetic forms of oxygen, probably with molecular ionO₂ ⁺ or with excited O₂ molecules, and this process is thermodynamicallyand kinetically favorable, however, a high energy consumption isrequired. Therefore, there is a need for an improved method of mercuryremoval from gas streams that overcomes the above problems.

The consideration of the emissions control problems concerning NO-to-NO₂conversion, SO₃ -to-SO₂ conversion, CO and organic compound removal,shows that OH and HO₂ radicals play important roles in pollution controlreactions in combustion gases. It would be of great economic andenvironmental benefit to find an additive to be injected into combustiongases and capable of producing additional OH and HO₂ radicals. Such aprocess combined with simultaneous NO₂ and SO₂ scrubbing would be a stepforward in the effective control of the main air pollutants. It wouldmake it easier to control multiple emissions and it would increase theutilization of sulfur containing fuels, both coal and oil. It would be arelated advancement to convert mercury vapor to less volatile mercuryoxide for its subsequent removal by ESP. If HO₂ radicals can rapidlyreact with mercury atoms, the additive which forms HO₂ radicals couldalso remove Hg vapor. Such additives and apparatus are disclosed andclaimed below.

SUMMARY OF THE INVENTION

In view of the present state of the art, it is an object of the presentinvention to provide a method for removing NO from combustion flue gasbefore it is emitted to the atmosphere.

It is another object of the present invention to quickly and efficientlyoxidize NO to more reactive NO₂ for its removal by different scrubbingmethods.

It is a further object of the present invention to convert NO to NO₂without forming other air pollutants.

It is also another object of the present invention to provide a methodfor removing SO₃ from gas streams in the form of SO₂.

It is also another object of the present invention to convert SO₃ to SO₂for its removal by different scrubbing methods.

It is a further object of the present invention to remove SO₃ from gasstreams to reduce corrosion of industrial equipment.

It is also another object of the present invention to remove SO₃ fromflue gas for increasing thermal efficiency of power plants by loweringthe SO₃ dewpoint and by enlargement of the air heater.

It is also another object of the invention to provide a method forpreventing the formation of sulfuric acid in combustion flue gas.

It an additional object of the present invention to remove hydrocarbonsfrom gas streams by oxidizing them to CO₂.

It is also another object of the present invention to reduce theconcentration of CO in combustion flue gas by oxidizing CO to CO₂.

It a further object of the present invention to reduce the concentrationof CO in gas streams during injection of methanol by substitution of apart of methanol with hydrogen peroxide.

It is also another object of the present invention to reduce the cost ofusing hydrogen peroxide by its partial substitution with methanolwithout exceeding required CO levels.

It is still another object of the present invention to provide a methodfor quickly and efficiently oxidizing mercury vapor in gas streamsbefore emitting into the atmosphere so as to be able to remove mercuryin the form of the less volatile HgO.

Additional objects and advantages of the present invention will beapparent from the description and claims which follow, or may be learnedby the practice of the invention.

In accordance with the foregoing objects and advantages, the presentinvention is a method for the removal of multiple air pollutants fromcombustion flue gas streams, such as the gas components including one ormore of NO, SO₃, CO, light hydrocarbons, and mercury vapor. The methodcomprises the step of contacting the gas components with hydrogenperoxide (H₂ O₂) in an amount such that the mole ratio of the hydrogenperoxide to the sum of any such NO, SO₃, CO, light hydrocarbons andmercury vapor contained in the combustion flue gas is in the range fromabout 0.5 to about 2.0. The hydrogen peroxide substantially converts theNO, SO₃, CO, light hydrocarbons, and mercury vapor in the combustionflue gas to NO₂, SO₂, CO₂, and HgO. In an alternative embodiment, amixture of hydrogen peroxide and methanol can be used in the abovemethod.

The method of the invention can be accomplished by injecting aconcentration of hydrogen peroxide or a H₂ O₂ /CH₃ OH mixture into theflue gas, and allowing the chemicals to react within the flue gas for asufficient time so as to convert one or more of NO, SO₃, CO, lighthydrocarbons, and mercury to NO₂, SO₂, CO₂, and HgO. The combustion fluegas can further comprise initial concentrations of carbon dioxide, waterand oxygen. Preferably, the chemicals are injected into the flue gas ata temperature within the range from about 650 K (377° C.) to about 1100K (827° C.). Temperatures that are significantly outside of this rangetend to greatly reduce the effectiveness of the process. The reactiontime of the H₂ O₂ or H₂ O₂ /CH₃ OH mixture should be in the range fromabout 0.01 to about 5 seconds, preferably from about 0.1 to about 2seconds. The NO₂, SO₂, and HgO can then be removed from the flue gas.The hydrogen peroxide or H₂ O₂ /CH₃ OH mixture can be injected into theflue gas upstream of a particulate control device.

Small additions of hydrogen peroxide, in either a gas or liquid form,have been found to remove NO, SO₃, hydrocarbons, CO, and mercury vaporfrom the flue gas. When injected into a flue gas stream, the hydrogenperoxide or H₂ O₂ /CH₃ OH mixture comprises less than about 1,000 ppm ofthe total gas stream, and preferably less than about 500 ppm of thetotal gas stream.

The mechanisms of air pollutants removal include different chainreactions. These processes occur when the reactions between the airpollutants and HO₂ or OH radicals exist:

    NO+HO.sub.2 →NO.sub.2 +OH                           (1)

    SO.sub.3 +HO.sub.2 →HSO.sub.3 +O.sub.2              ( 2)

    CH.sub.4 +OH→H.sub.2 O+CH.sub.3                     ( 5)

    CO+OH→CO.sub.2 +H                                   (4)

    Hg+HO.sub.2 →HgO+OH                                 (6)

The above reactions occur when hydrogen peroxide or a H₂ O₂ /CH₃ OHmixture is injected into combustion flue gas at temperatures of about650 K (377° C.) and higher. Initially, the hydrogen peroxide dissociatesinto two hydroxyl radicals by the reaction:

    H.sub.2 O.sub.2 +M→OH+OH+M                          (7)

This reaction thus serves as a chain initiation step. The OH radicalsthat are formed then react either with air pollutants, such as CO andhydrocarbons (reactions 4 and 5), or with another molecule of H₂ O₂ viaa rapid chain propagating reaction of:

    OH+H.sub.2 O.sub.2 →HO.sub.2 +H.sub.2 O             (8)

Once this reaction occurs, the HO₂ radicals return OH radicals throughthe reactions with NO (reaction 1), SO₃ (reaction 2), and Hg (reaction6) to form NO₂, HSO₃, and HgO as stated above. The HSO₃ moleculesdissociate under these conditions to form SO₂ and return OH radicals:

    HSO.sub.3 +M→SO.sub.2 +OH+M                         (3)

As discussed above, the amount of H₂ O₂ or a H₂ O₂ /CH₃ OH mixtureinjected into gas streams should be in the approximate mole ratio fromabout 0.8 to about 2.0 with respect to the sum of the air pollutants:NO, SO₃, CO, hydrocarbons, and mercury vapor. In most applications, thepreferable mole ratio will be from about 0.9 to about 1.5.

BRIEF DESCRIPTION OF THE DRAWINGS

In order that the manner in which the above-recited and other advantagesand objects of the invention are obtained, a more particular descriptionof the invention briefly described above will be rendered by referenceto the appended drawings. Understanding that these drawings are not tobe considered limiting of its scope, the invention will be described andexplained with additional specificity and detail through the use of theaccompanying drawings in which:

FIGS. 1A and 1B are graphs representing the experimental (a) andmodeling (b) data for NO-to-NO₂ conversion by H₂ O₂ injection. t_(r)(reaction time)=1.0-2.0 s; mixture (1): 100 ppm NO, (160÷220) ppm H₂ O₂,4.2% O₂, 5.4% H₂ O, balance N₂ ; mixture (2): 100 ppm NO, (90÷120) ppmH₂ O₂, 4.4% O₂, 1.9% H₂ O, balance N₂.

FIGS. 2A and 2B are graphs representing experimental (a) and modeling(b) data for SO₃ -to-SO₂ conversion by H₂ O₂ injection. t_(r) =1.0-1.6s; mixture (1): 100 ppm SO₃, (160÷220) ppm H₂ O₂, 4.2% O₂, 5.4% H₂ O,balance N₂ ; mixtures (2-5) are the same but with different amounts ofH₂ O₂ : (2) without H₂ O₂, (3) 100 ppm H₂ O₂, (4) 200 ppm H₂ O₂, (5) 500ppm H₂ O₂.

FIGS. 3A and 3B are graphs representing experimental (a) and modeling(b) data for CO oxidation by H₂ O₂ injection. t_(r) =1.0-1.5 s; mixture(1): 90 ppm CO, (160÷220) ppm H₂ O₂, 4.2% O₂, 5.4% H₂ O, balance N₂ ;mixture (2): 90 ppm CO, 4.2% O₂, 5.4% H₂ O, balance N₂ ; mixture (3): 90ppm CO, 4.2% O₂, balance N₂ (10 ppm H₂ O in modeling).

FIGS. 4A and 4B are graphs representing experimental (a) and modeling(b) data for CH₄ oxidation by H₂ O₂ injection. t_(r) =1.0-1.8 s; mixture(1): 90 ppm CH₄, (160÷220) ppm H₂ O₂, 4.2% O₂, 5.4% H₂ O, balance N₂ ;mixture (2): 90 ppm CH₄, (90÷120) ppm H₂ O₂, 4.4% O₂, 1.9% H₂ O, balanceN₂ ; mixture (3): the same as (1) but without H₂ O₂.

FIGS. 5A and 5B are graphs representing pilot-scale measurements of NOand CO concentrations after injection of H₂ O₂ (solid curves) andmethanol (dash curves). Additive!/ NO!=1.5, CO!₀ =24 ppm, (a) NO!₀ =400ppm, (b) NO!₀ =200 ppm.

FIGS. 6A and 6B are graphs representing pilot-scale measurements of NOand CO concentrations after injection of H₂ O₂ /CH₃ OH mixtures. ( H₂ O₂!+ CH₃ OH!)/ NO!₀ =1.5, NO!₀ =70 ppm, CO!₀ =30 ppm, (a) temperaturewindows for various H₂ O₂ /CH₃ OH mixtures, (b) NO and CO concentrationsat 800 K (527° C.) (at H₂ O₂ !=0 data are shown for 866 K (593° C.)).

FIG. 7 is a graph representing the temperature dependence of vaporpressure over condensed phase for most important mercury compounds.These data are taken from Hall et al., Mercury Chemistry in SimulatedFlue Gases Related to Waste Incineration Conditions, Environ. Sci.Technol., 1990, V. 24, pp. 108-111, which is incorporated herein byreference.

FIG. 8 is a graph representing a kinetic modeling of mercury removal byhydrogen peroxide injection at a temperature of 800 K (527° C.).

DETAILED DESCRIPTION OF THE INVENTION

As will be appreciated by consideration of the following description aswell as the accompanying Figures, the inventive concepts of the presentinvention may be embodied in different forms. The presently preferredembodiment described herein represents the presently preferred best modefor carrying out the invention. Nevertheless, many embodiments, orvariations of the preferred embodiment, other than those specificallydetailed herein, may be used to carry out the inventive conceptsdescribed in the claims appended hereto.

According to the present invention, a novel method for selectivenon-catalytic reduction of multiple air pollutants in gas streamsinvolves different homogeneous gas-phase chain reactions. The method ofthe present invention comprises injecting hydrogen peroxide or a H₂ O₂/CH₃ OH mixture into a combustion flue gas stream in order to convertunreactive NO to much more reactive NO₂, corrosive SO₃ to SO₂, CO andlight hydrocarbons to CO₂, and mercury vapor to the less volatilemercury oxide. The method of the invention can further comprise the stepof removing the NO₂, SO₂, and HgO from the flue gas. The use of hydrogenperoxide to provide OH and HO₂ radicals allows removal of multiple airpollutants in the form of more stable or more reactive compounds, whichare then easier to remove by other methods.

Small additions of hydrogen peroxide, which can be either in a gas orliquid form, have been found to remove NO, SO₃, hydrocarbons, CO, andmercury vapor from the combustion flue gas. When injected into a fluegas stream, the hydrogen peroxide or H₂ O₂ /CH₃ OH mixture comprisesless than about 1,000 ppm of the total gas stream, and preferably lessthan about 500 ppm of the total gas stream. The initial concentration ofNO in the flue gas is up to about 1000 ppm, while the initialconcentration of SO₃ in the flue gas is up to about 100 ppm. The initialconcentration of CO in the flue gas is up to about 500 ppm, while theinitial concentration of light hydrocarbons in the flue gas is up toabout 1000 ppm. The initial concentration of Hg in the flue gas is lessthan about 1 ppm. The combustion flue gas can further include initialconcentrations of carbon dioxide, water and oxygen.

The present invention provides a quick and efficient method of removingone or more of NO, SO₃, CO, light hydrocarbons, and mercury emitted fromdifferent sources such that these air pollutants are not emitted intothe atmosphere in dangerously high concentrations. For example, thepresent invention can be used to remove the air pollutants in flue gasesemanating from boilers, furnaces, incinerators, stationary engines, andother systems connected with combustion of fossil fuels. The process canalso be used for multiple air pollutants control in industrial gasescontaining the air pollutants.

The amount of hydrogen peroxide and methanol which must be used isdirectly related to the total amount of the air pollutants present inthe gas streams. Accordingly, it is important to the present inventionto provide enough hydrogen peroxide and methanol to accomplishsubstantial reduction of the air pollutants, but not enough hydrogenperoxide to cause its unnecessary consumption and not enough methanol toexceed the required limits for CO emissions. The total amount ofhydrogen peroxide and methanol used in combination with the gas streamswill generally be in the mole ratio of from about 0.5 to about 2.0, butin most applications from about 0.9 to about 1.5 of the total airpollutants. NO concentrations are usually much more easily measured thanSO₃ concentrations and Hg concentrations. The amount of NO in combustionflue gases is much larger than the amount of SO₃ and several orders ofmagnitude larger than the amount of Hg vapor. Hence, in most situations,it is both acceptably accurate and more convenient to control the amountof the additives used with relationship to the relatively total amountof NO, CO, and hydrocarbons without addition of SO₃ and Hgconcentrations.

The temperature at which the chemicals are introduced into the gasstream must be carefully controlled in order to produce the desiredresults. It is presently preferred that the temperature be within therange of about 650 K (377° C.) to about 1100 K (827° C.). When thetemperature is significantly below 650 K (377° C.), it has been foundthat the hydrogen peroxide will not form enough OH radicals to supportthe chain reactions and the air pollutants will not be removed.

This is also true for excessively high temperatures. When thetemperature is higher than about 1100 K (827° C.), the hydrogen peroxidedissociates faster than the reactions with air pollutants can occur. Onthe other hand, at higher temperatures, the concentration ofhydrocarbons and CO can be reduced by the oxidation reactions withoxygen. At higher temperatures, NO and SO₃ concentrations can be reducedby the reactions with methanol.

It is presently preferred that reaction times be held within the rangefrom about 0.01 to about 5 seconds, and most preferably from about 0.1to about 2 seconds. The preferred reaction time will vary with thetemperature at which the reaction occurs.

According to the present invention, hydrogen peroxide can be injectedpreferably in the form of an aqueous solution having a concentration ofabout 1% to 50%, and more preferably from about 10 to 30%. Hydrogenperoxide can be also injected as a mixture of H₂ O₂ solution andmethanol. The use of H₂ O₂ and methanol mixtures is presently preferredbecause methanol is very low in cost. The methanol to H₂ O₂ ratio shouldbe as high as possible to reduce the cost of the additive, but tosatisfy CO emission requirements.

In one embodiment of the present invention, aqueous H₂ O₂ or a solutionof H₂ O₂ and methanol is atomized by a jet of gas and then propelledinto the combustion flue gas by the jet of gas. The vaporization of theliquid droplets is a process requiring a finite time. Hence, contactingthe chemicals with the flue gas occurs not immediately upon injectionbut only after some delay.

When the present invention is used in combination with SNCR NO_(x)reduction technologies, the hydrogen peroxide or H₂ O₂ /CH₃ OH mixtureand the combustion flue gas contact downstream of the zone in which thecombustion effluents are contacted with a SNCR reducing agent. Thehydrogen peroxide or H₂ O₂ /CH₃ OH mixture can also be injected into theflue gas upstream of a particulate control device.

It is well known that complex chemical reactions occur by a series ofelementary reactions, and that if one knows the rate constants of suchsteps a theoretical kinetic mechanism can be developed and used tocalculate the behavior of the reaction under any set of conditions.Mechanisms describing the decomposition of H₂ O₂, oxidation of lighthydrocarbons and their oxygenates, such as methanol, the interaction ofNO and other nitrogen compounds at high temperatures, and theinteraction of sulfur compounds have been developed in differentchemical kinetics studies. Assembling these mechanisms produces a totalkinetic mechanism to describe the chemical reactions of this invention.Such a mechanism is assumed in the examples set forth below.

The temperature dependence of vapor pressure over condensed phase formost important Hg compounds is presented in FIG. 7. At temperatureslower than 170° C., mercury vapor is the most volatile and dominatingconstituent in the gas phase of the compounds listed. A mercury vaporconcentration in flue gas of about 10 ppm and less, is well below thesaturation pressure. Therefore, in the absence of a chemical or physicalmercury control process, mercury vapor will be emitted to theatmosphere.

Mercury oxide is a less volatile compound as compared to mercury. At150° C., the saturation HgO pressure is about 10 ppb. At 100° C., thesaturation HgO pressure is about 100 ppt. At 50° C., the saturation HgOpressure is about 0.1 ppt. If the HgO concentration exceeds theselevels, HgO will form solid particles which can be removed inconventional fashion, such as by ESP. Therefore, depending on workingtemperatures of particulate control devices and on initial mercuryconcentrations, conversion of elemental mercury into HgO is a possibleway for mercury removal.

The use of hydrogen peroxide in the present invention has manyadvantages. If properly stored, hydrogen peroxide solutions in water arevery stable. The use of hydrogen peroxide does not pose anyenvironmental problems since hydrogen peroxide is not itself a source ofpollution, and the only reaction by-products are water and oxygen.Therefore, hydrogen peroxide can be used safely in the presentinvention.

Once the H₂ O₂ is injected into combustion flue gas, the hydrogenperoxide dissociates into two hydroxyl radicals by the followingreaction:

    H.sub.2 O.sub.2 +M→2OH+M                            (7)

The hydroxyl radicals formed have several reaction routes. First, theycan react with H₂ O₂ molecules to form HO₂ radicals:

    OH+H.sub.2 O.sub.2 →H.sub.2 O+HO.sub.2              (8)

Second, OH radicals interact with carbon-containing compounds, such asCO, CH₄, CH₃ OH, and other organics:

    OH+CO→CO.sub.2 +H                                   (4)

    OH+CH.sub.4 →H.sub.2 O+CH.sub.3                     (5)

    OH+CH.sub.3 OH→H.sub.2 O+CH.sub.2 OH                (9)

followed by oxidation reactions:

    H+O.sub.2 →OH+O                                     (10)

    CH.sub.3 +O.sub.2 →CH.sub.2 O+OH                    (11)

    CH.sub.2 OH+O.sub.2 →CH.sub.2 O+HO.sub.2            (12)

which increases concentrations of active species. Under thesehigh-temperature conditions, CH₂ O formed is converted to CO, CO₂, andH₂ O via a CH₂ O--O₂ chain reaction. Total stoichiometry of the CH₄ /O₂reaction is well known:

    CH.sub.4 +2O.sub.2 =CO.sub.2 +2H.sub.2 O                   (13)

and it is promoted in the presence of OH radicals. As known from theliterature, H₂ O₂ enhances oxidation of different organic compounds dueto the chain processes involving OH and other active species. Forinstance, it was found by Cooper et al., Enhancement of Organic VaporIncineration Using Hydrogen Peroxide, J. Hazard. Mat., 27, 273-285,1991, which is incorporated herein by reference, that injection of H₂ O₂in dilute air mixtures of heptane and isopropanol increases the rate oftheir destruction at T=910-1073 K (637°-800° C.) and t_(r) =0.26-0.94 s.

Third, the OH radicals participate in chain termination steps, such asOH+HO₂ →H₂ O+O₂, OH+OH+M→H₂ O₂ +M, etc.

Thus, the reaction (7) serves as a chain initiation step, and once theOH radicals are formed, they can then react with H₂ O₂ through the chainpropagation reaction step (8). Once this reaction occurs, the HO₂radicals return OH radicals through the reactions with air pollutants.Both OH and HO₂ radicals play an important role in pollutants reduction.

Comparison of modeling and experimental results (FIGS. 1A and 1B throughFIGS. 4A and 4B) is complicated because of H₂ O₂ surface reactionsalthough the heterogeneous H₂ O₂ decomposition is taken into account inmodeling. Nevertheless, modeling at least qualitatively describes allsubstantial features of NO, SO₃, CO and CH₄ reduction except the minimumof the SO₂ curve (FIG. 2B). Most likely, this minimum is not a result ofhomogeneous chemical reactions that can be checked in future tests.

Both modeling and experimental results show that NO is not converted toNO₂ in the absence of H₂ O₂, but SO₃, CO, and CH₄ are converted to SO₂and CO₂ at higher temperatures even without H₂ O₂ addition (curves 2, 3and 3 in FIGS. 2B, 3A and 3B, 4A and 4B, respectively). However, innon-ideal practical combustion systems all these pollutants (SO₃ andcarbon-containing compounds) are present in flue gas, and H₂ O₂injection will reduce their concentrations.

The pilot-scale tests on NO-to-NO₂ conversion (FIGS. 5A, 5B and 6A, 6B)confirmed that performance of H₂ O₂ injection is higher when theinfluence of surface reactions is lower. Very low initial CO levels inthe pilot-scale tests (24-30 ppm) did not allow to demonstrate COreduction. Maximum CO reduction in the laboratory tests (FIG. 3A) wasonly about 20% (2-3 ppm for pilot experiments), which is close to the COdetection limit.

The position of the H₂ O₂ temperature window is approximately in thesame range for all pollutants: between 650-1100 K (377°-827° C.), withmaximum performance between 700-1000 K (427°-727° C.). The position ofthe H₂ O₂ temperature window is defined by the chemical nature of H₂ O₂reactions. At temperatures lower than 650 K (377° C.), the homogeneousH₂ O₂ decomposition is very slow and OH and HO₂ radicals are not formed.At temperatures higher than 1100 K (827° C.), concentrations of allradicals in the system become very high, and the rate of recombinationreactions which are quadratic on radical concentration prevails in therate of their reactions with molecules. As a result, H₂ O₂ dissociatesinto hydroxyl radicals very rapidly, and the radicals disappear in therecombination processes. An important factor is also the decompositionof HO₂ radicals at temperatures higher than 1000 K. Thus, H₂ O₂ isactive only in the temperature range of about 650-1100 K (377°-827° C.).

The most important chain reactions that are responsible for reduction ofair pollutants are as follows:

    ______________________________________                                        NO removal:                                                                   OH + H.sub.2 O.sub.2 → H.sub.2 O + HO.sub.2                                                   (8)                                                    HO.sub.2 + NO → NO.sub.2 + OH                                                                 (1) (chain reaction)                                   SO.sub.3 removal:                                                             OH + H.sub.2 O.sub.2 → H.sub.2 O + HO.sub.2                                                   (8)                                                    HO.sub.2 + SO.sub.3 → HSO.sub.3 + O.sub.2                                                     (2)                                                    HSO.sub.3 + M → SO.sub.2 + OH + M                                                             (3) (chain reaction)                                   CO reduction:                                                                 OH + CO → CO.sub.2 + H                                                                        (4)                                                    H + O.sub.2 → OH + O                                                                          (10) (chain reaction)                                  CH.sub.4 reduction:                                                           the reaction CH.sub.4 + 2O.sub.2 = CO.sub.2 + 2H.sub.2 O                                             (13) (chain reaction)                                  is promoted in the presence of OH radicals.                                   Hg removal:                                                                   OH + H.sub.2 O.sub.2 → H.sub.2 O + HO.sub.2                                                   (8)                                                    HO.sub.2 + Hg → HgO + OH                                                                      (6) (chain reaction)                                   ______________________________________                                    

Thus, five various chain reactions are involved, and the single reagentof hydrogen peroxide can provide the reduction of multiple airpollutants.

The mechanism of mercury vapor removal includes a chain reaction betweenmercury and hydrogen peroxide. This process occurs when the chainpropagation elementary reaction between mercury atoms and HO₂ radicalsexists. The reaction that occurs is believed to be the second chainpropagation reaction:

    HO.sub.2 +Hg→HgO+OH                                 (6)

and once the chain reaction has occurred, much of the Hg is converted toHgO. The HgO can then be precipitated out and removed safely, withoutsending dangerous Hg vapor into the atmosphere. Thus, when the reaction(6) takes place, mercury is effectively removed from gas streams. It isbelieved that this reaction may occur with a rate constant close to thecollision frequency with the rate constant of about 10¹³.5 cc/mol.s.Hydrogen peroxide is used to provide oxygen atoms to be bonded withmercury atoms. According to the method of the present invention, thehydrogen peroxide is injected into the flue gas at temperatures of about650 K (377° C.) and higher. FIG. 8 is a graph representing a kineticmodeling of mercury removal by hydrogen peroxide injection at atemperature of 800 K (527° C.). It has been found that at temperatureslower than 650 K (377° C.), the H₂ O₂ is stable and does not produce thedesired radicals within a reaction time of several seconds. Conversely,at temperatures too high (above about 1100 K (827° C.)), the H₂ O₂dissociates too fast.

The methods of the present invention allow for the removal of SO₃ fromgas streams to reduce corrosion of industrial equipment, and to increasethermal efficiency of power plants by lowering the SO₃ dewpoint and byenlargement of the air heater. The methods of the invention also preventthe formation of sulfuric acid in combustion flue gas.

EXAMPLES

The following examples are given to illustrate the process of thepresent invention, but the examples are not intended to limit the scopeof the present invention.

Chemical kinetics calculations can describe the reactions of hydrogenperoxide with different air pollutants. Kinetic modeling was used todescribe and to test the reactions that take place in the method of theinvention and to compare the performance of H₂ O₂ with experimentaldata. Table I below presents the chemical reaction mechanism that wasused for kinetic modeling. The mechanism includes usually acceptedreactions of H₂ --O₂ interaction, SO₃ --SO₂ reactions, NO_(x) formationand destruction, and CH₄ oxidation reactions. Rate constants (k_(i)) forNO_(x) --H₂ --O₂ --CH₄ reactions were taken from Miller, J. A. andBowman, C. T., Mechanism and Modeling of Nitrogen Chemistry inCombustion, Progr. Energy and Combust. Sci., v. 15, pp. 287-338 (1989),which is incorporated herein by reference. Other rate constants weretaken from the NIST Chemical Kinetics Database, Version 5 (1993) whichis incorporated herein by reference. The rate constant for mercuryinteraction with HO₂ radicals was varied in modeling. The CHEMKIN-IIkinetic program developed by Sandia National Laboratories was used formodeling. Reverse reactions were also taken into account.

                  TABLE I                                                         ______________________________________                                        Chemical mechanism which was used for modeling                                k.sub.i = A · T.sup.n exp (-E/RT) (kcal, cm, mol, s)                 ELEMENTS: H, O, N, C, S.                                                      SPECIES: CH.sub.4, CH.sub.3, CH.sub.2, CH, CH.sub.2 O, HCO, C.sub.2 H,        CO.sub.2, CO, H.sub.2, H, O.sub.2,                                            O, OH, HO.sub.2, H.sub.2 O.sub.2, H.sub.2 O, C, C.sub.2 H.sub.4, C.sub.2      H.sub.3, C.sub.2 H.sub.5,                                                     C.sub.2 H.sub.6, C.sub.2 H.sub.2, CH.sub.3 O, CH.sub.2 OH, N.sub.2, NO,       N, NH, NH.sub.2, HNO, HCN,                                                    NCO, CN, N.sub.2 O, NNH, NH.sub.3, N.sub.2 H.sub.2, C.sub.2 N.sub.2,          NO.sub.2, HNO.sub.2, HOCN,                                                    HCNO, HNCO, SO.sub.2, SO.sub.3, HSO.sub.3, O.sub.3, CH.sub.3 OH, H.sub.2      O.sub.2 (wall).                                                               REACTIONS           A         n      E                                        ______________________________________                                        CH.sub.3 + CH.sub.3 = C.sub.2 H.sub.6                                                             9.03E16   -1.2   654                                      CH.sub.3 + H + M = CH.sub.4 + M                                                                   8.0E26    -3.0   0                                        H.sub.2 /2/ CO/2/ CO.sub.2 /3/ H.sub.2 O/5/                                   CH.sub.4 + O.sub.2 = CH.sub.3 + HO.sub.2                                                          7.9E13    0.0    56000                                    CH.sub.4 + H = CH.sub.3 + H.sub.2                                                                 2.2E4     3.0    8750                                     CH.sub.4 + OH = CH.sub.3 + H.sub.2 O                                                              1.6E6     2.1    2460                                     CH.sub.4 + O = CH.sub.3 + OH                                                                      1.02E9    1.5    8604                                     CH.sub.4 + HO.sub.2 = CH.sub.3 + H.sub.2 O.sub.2                                                  1.8E11    0.0    18700                                    CH.sub.3 + O.sub.2 = CH.sub.3 O + O                                                               2.05E18   -1.57  29229                                    CH.sub.3 + HO.sub.2 = CH.sub.3 O + OH                                                             2.0E13    0.0    0                                        CH.sub.3 + O = CH.sub.2 O + H                                                                     8.00E13   0.0    0                                        CH.sub.2 OH + H = CH.sub.3 + OH                                                                   1.0E14    0.0    0                                        CH.sub.3 O + H = CH.sub.3 + OH                                                                    1.0E14    0.0    0                                        CH.sub.3 + OH = CH.sub.2 + H.sub.2 O                                                              7.5E6     2.0    5000                                     CH.sub.3 + H = CH.sub.2 + H.sub.2                                                                 9.0E13    0.0    15100                                    CH.sub.3 O + M = CH.sub.2 O + H + M                                                               1.0E14    0.0    25000                                    CH.sub.2 OH + M + CH.sub.2 O + H + M                                                              1.0E14    0.0    25000                                    CH.sub.3 O + H = CH.sub.2 O + H.sub.2                                                             2.0E13    0.0    0                                        CH.sub.2 OH + H = CH.sub.2 O + H.sub.2                                                            2.0E13    0.0    0                                        CH.sub.3 O + OH = CH.sub.2 O + H.sub.2 O                                                          1.0E13    0.0    0                                        CH.sub.2 OH + OH = CH.sub.2 O + H.sub.2 O                                                         1.0E13    0.0    0                                        CH.sub.3 O + O = CH.sub.2 O + OH                                                                  1.0E13    0.0    0                                        CH.sub.2 OH + O = CH.sub.2 O + OH                                                                 1.0E13    0.0    0                                        CH.sub.3 O + O.sub.2 = CH.sub.2 O + HO.sub.2                                                      6.3E10    0.0    2600                                     CH.sub.2 OH + O.sub.2 = CH.sub.2 O + HO.sub.2                                                     1.48E13   0.0    1500                                     CH.sub.2 + H = CH + H.sub.2                                                                       1.0E18    -1.56  0                                        CH.sub.2 + OH = CH.sub.2 O + H                                                                    2.5E13    0.0    0                                        CH.sub.2 + OH = CH + H.sub.2 O                                                                    1.13E7    2.0    3000                                     CH + O.sub.2 = HCO + O                                                                            3.3E13    0.00   0                                        CH + O = CO + H     5.7E13    0.0    0                                        CH + OH =HCO + H    3.0E13    0.0    0                                        CH + CO.sub.2 = HCO + CO                                                                          3.4E12    0.0    690                                      CH + H = C + H.sub.2                                                                              1.5E14    0.0    0                                        CH + H.sub.2 O = CH.sub.2 O + H                                                                   1.17E15   -0.75  0                                        CH + CH.sub.2 = C.sub.2 H.sub.2 + H                                                               4.0E13    0.0    0                                        CH + CH.sub.3 = C.sub.2 H.sub.3 + H                                                               3.0E13    0.0    0                                        CH + CH.sub.4 =C.sub.2 H.sub.4 + H                                                                6.0E13    0.0    0                                        C + O.sub.2 = CO + O                                                                              2.0E13    0.0    0                                        C + OH = CO + H     5.0E13    0.0    0                                        C + CH.sub.3 = C.sub.2 H.sub.2 + H                                                                5.0E13    0.0    0                                        C + CH.sub.2 = C.sub.2 H + H                                                                      5.0E13    0.0    0                                        CH.sub.2 + CO.sub.2 = CH.sub.2 O + CO                                                             1.1E11    0.0    1000                                     CH.sub.2 + O = CO + H + H                                                                         5.0E13    0.0    0                                        CH.sub.2 + O = CO + H.sub.2                                                                       3.0E13    0.0    0                                        CH.sub.2 +O .sub.2 = CO.sub.2 + H + H                                                             1.6E12    0.0    1000                                     CH.sub.2 +O .sub.2 = CH.sub.2 O + O                                                               5.0E13    0.0    9000                                     CH.sub.2 +O .sub.2 = CO.sub.2 + H.sub.2                                                           6.9E11    0.0    500                                      CH.sub.2 +O .sub.2 = CO + H.sub.2 O                                                               1.9E10    0.0    -1000                                    CH.sub.2 +O .sub.2 = CO + OH + H                                                                  8.6E10    0.0    -500                                     CH.sub.2 +O .sub.2 = HCO + OH                                                                     4.3E10    0.0    -500                                     CH.sub.2 O + OH = OHCO + H.sub.2 O                                                                3.43E9    1.18   -447                                     CH.sub.2 O + H = HCO + H.sub.2                                                                    2.19E8    1.77   3000                                     CH.sub.2 O + M = HCO + H + M                                                                      3.31E16   0.0    81000                                    CH.sub.2 O + O = HCO + OH                                                                         1.8E13    0.0    3080                                     HCO + OH = H.sub.2 O + CO                                                                         1.0E14    0.0    0                                        HCO + M = H + CO + M                                                                              2.5E14    0.0    16802                                    CO/1.9/H.sub.2 /1.9/CH.sub.4 /2.8/CO.sub.2 /3./H.sub.2 O/5./                  HCO + H = CO + H.sub.2                                                                            1.19E13   0.25   0                                        HCO + O = CO + OH   3.0E13    0.0    0                                        HCO + O = CO.sub.2 + H                                                                            3.0E13    0.0    0                                        HCO + O.sub.2 = HO.sub.2 + CO                                                                     3.3E13    -0.4   0                                        CO + O + M = CO.sub.2 + M                                                                         6.17E14   0.0    3000                                     CO + OH = CO.sub.2 + H                                                                            1.51E7    1.3    -758                                     CO + O.sub.2 = CO.sub.2 + 0                                                                       1.6E13    0.0    41000                                    HO.sub.2 + CO = CO.sub.2 + OH                                                                     5.8E13    0.0    22934                                    C.sub.2 H.sub.6 + CH.sub.3 =  C.sub.2 H.sub.5 + CH.sub.4                                          5.5E - 1  4.0    8300                                     C.sub.2 H.sub.6 + H = C.sub.2 H.sub.5 + H.sub.2                                                   5.4E2     3.5    5210                                     C.sub.2 H.sub.6 + O = C.sub.2 H.sub.5 + OH                                                        3.0E7     2.0    5115                                     C.sub.2 H.sub.6 + OH = C.sub.2 H.sub.5 + H.sub.2 O                                                8.7E9     1.05   1810                                     C.sub.2 H.sub.4 + H = C.sub.2 H.sub.3 + H.sub.2                                                   1.1E14    0.0    8500                                     C.sub.2 H.sub.4 + O = CH.sub.3 + HCO                                                              1.6E9     1.2    746                                      C.sub.2 H.sub.4 + OH = C.sub.2 H.sub.3 + H.sub.2 O                                                2.02E13   0.0    5955                                     CH.sub.2 + CH.sub.3 = C.sub.2 H.sub.4 + H                                                         3.0E13    0.0    0                                        H + C.sub.2 H.sub.4 = C.sub.2 H.sub.5                                                             2.21E13   0.0    2066                                     C.sub.2 H.sub.5 + H = CH.sub.3 + CH.sub.3                                                         1.0E14    0.0    0                                        C.sub.2 H.sub.5 + O.sub.2 = C.sub.2 H.sub.4 + HO.sub.2                                            8.4E11    0.0    3875                                     C.sub.2 H.sub.2 + O = CH.sub.2 + CO                                                               1.02E7    2.0    1900                                     H.sub.2 + C.sub.2 H = C.sub.2 H.sub.2 + H                                                         4.09E5    2.39   864                                      H + C.sub.2 H.sub.2 = C.sub.2 H.sub.3                                                             5.54E12   0.0    2410                                     C.sub.2 H.sub.3 + H = C.sub.2 H.sub.2 + H.sub.2                                                   4.0E13    0.0    0                                        C.sub.2 H.sub.3 + O.sub.2 = CH.sub.2 O = HCO                                                      4.0E12    0.0    -250                                     C.sub.2 H.sub.3 + OH = C.sub.2 H.sub.2 + H.sub.2 O                                                5.0E12    0.0    0                                        C.sub.2 H.sub.3 + CH.sub.2 = C.sub.2 H.sub.2 + CH.sub.3                                           3.0E13    0.0    0                                        C.sub.2 H.sub.3 + C.sub.2 H = C.sub.2 H.sub.2 + C.sub.2 H.sub.2                                   3.0E13    0.0    0                                        C.sub.2 H.sub.3 + CH = CH.sub.2 + C.sub.2 H.sub.2                                                 5.0E13    0.0    0                                        C.sub.2 H.sub.2 + OH = C.sub.2 H + H.sub.2 O                                                      3.37E7    2.0    14000                                    C.sub.2 H.sub.2 + OH = CH.sub.3 + CO                                                              4.83E - 4 4.0    -2000                                    C.sub.2 H.sub.2 + O = C.sub.2 H + OH                                                              3.16E15   -0.6   15000                                    C.sub.2 H + O.sub.2 = HCO + CO                                                                    5.0E13    0.0    1500                                     C.sub.2 H + O = CH + CO                                                                           5.0E13    0.0    0                                        CH.sub.2 + CH.sub.2 = C.sub.2 H.sub.2 + H.sub.2                                                   4.0E13    0.0    0                                        C.sub.2 H.sub.2 + M = C.sub.2 H + H + M                                                           4.2E16    0.0    107000                                   C.sub.2 H.sub.4 + M = C.sub.2 H.sub.2 + H.sub.2 + M                                               1.5E15    0.0    55800                                    C.sub.2 H.sub.4 + M = C.sub.2 H.sub.3 + H + M                                                     1.4E15    0.0    82360                                    H.sub.2 + O.sub.2 = 2OH                                                                           1.7E13    0.0    47780                                    OH + H.sub.2 = H.sub.2 O + H                                                                      1.17E9    1.3    3626                                     O + OH = O.sub.2 + H                                                                              4.0E14    -0.5   0                                        O + H.sub.2 = OH + H                                                                              5.06E4    2.67   6290                                     H + O.sub.2 + M = HO.sub.2 + M                                                                    3.61E17   -0.72  0                                        H.sub.2 O/18.6/CO.sub.2 /4.2/H.sub.2 /2.9/CO/2.1/N.sub.2 /1.3/                OH + HO.sub.2 = H.sub.2 O + O.sub.2                                                               2.0E13    0.0    0                                        H + HO.sub.2 = 2OH  1.4E14    0.0    1073                                     O + HO.sub.2 = O.sub.2 + OH                                                                       1.4E13    0.0    1073                                     2OH = O + H.sub.2 O 6.0E8     1.3    0                                        H + H + M = H.sub.2 + M                                                                           1.0E18    -1.0   0                                        H.sub.2 /0.0/H.sub.2 O/0.0/CO.sub.2 /0.0/                                     H + H + H.sub.2 = H.sub.2 + H.sub.2                                                               9.2E16    -0.6   0                                        H + H + H.sub.2 O = H.sub.2 + H.sub.2 O                                                           6.0E19    -1.25  0                                        H + H + CO.sub.2 = H.sub.2 + CO.sub.2                                                             5.49E20   -2.0   0                                        H + OH + M = H.sub.2 O + M                                                                        1.6E22    -2.0   0                                        H.sub.2 O/5/                                                                  H + O + M = OH + M  6.2E16    -0.6   0                                        H.sub.2 O/5/                                                                  O + O + M = O.sub.2 + M                                                                           1.89E13   0.0    -1788                                    H + HO.sub.2 = H.sub.2 + O.sub.2                                                                  1.25E13   0.0    0                                        HO.sub.2 + HO.sub.2 = H.sub.2 O.sub. 2 + O.sub.2                                                  2.0E12    0.0    0                                        H.sub.2 O.sub.2 + M = OH + OH + M                                                                 1.3E17    0.0    45500                                    H.sub.2 O.sub.2 + H = HO.sub.2 + H.sub.2                                                          1.6E12    0.0    3800                                     H.sub.2 O.sub.2 + OH = H.sub.2 O + HO.sub.2                                                       1.0E13    0.0    1800                                     CH + N.sub.2 = HCN + N                                                                            3.0E11    0.0    13600                                    CN + N = C + N.sub.2                                                                              1.04E15   -0.5   0                                        CH.sub.2 + N.sub.2 = HCN + NH                                                                     1.0E13    0.0    74000                                    C + NO = CN + O     6.6E13    0.0    0                                        CH + NO = HCN + O   1.1E14    0.0    0                                        CH.sub.2 + NO = HCNO + H                                                                          1.39E12   0.0    -1100                                    CH.sub.3 + NO = HCN + H.sub.2 O                                                                   1.0E11    0.0    15000                                    HCNO + H = HCN + OH 5.0E13    0.0    12000                                    CH.sub.2 + N = HCN + H                                                                            1.0E14    0.0    0                                        CH + N = CN + H     1.3E13    0.0    0                                        CO.sub.2 + N = NO + CO                                                                            1.9E11    0.0    3400                                     C.sub.2 H.sub.3 + N = HCN + CH.sub.2                                                              2.0E13    0.0    0                                        HCN + OH = CN + H.sub.2 O                                                                         1.45E13   0.0    10929                                    OH + HCN = HOCN + H 5.85E4    2.4    12500                                    OH + HCN = HNCO + H 1.98E - 4.0                                                                             1000                                            OH + HCN = NH.sub.2 + CO                                                                          7.83E - 4 4.0    4000                                     HOCN + H = HNCO + H 1.0E13    0.0    0                                        HCN + O = NCO + H   1.38E4    2.64   4980                                     HCN + O = NH + CO   3.45E3    2.64   4980                                     HCN + O = CN + OH   2.7E9     1.58   26600                                    CN + H.sub.2 = HCN + H                                                                            2.95E5    2.45   2237                                     CN + O = CO + N     1.8E13    0.0    0                                        CN + O.sub.2 = NCO + O                                                                            5.6E12    0.0    0                                        CN + OH = NCO + H   6.0E13    0.0    0                                        CN + HCN = C.sub.2 N.sub.2 + H                                                                    2.0E13    0.0    0                                        CN + NO.sub.2 = NCO + NO                                                                          3.0E13    0.0    0                                        CN + N.sub.2 O = NCO + N.sub.2                                                                    1.0E13    0.0    0                                        C.sub.2 N.sub.2 + 0 = NCO + CN                                                                    4.57E12   0.0    8880                                     C.sub.2 N.sub.2 + OH = HOCN + CN                                                                  1.86E11   0.0    2900                                     HO.sub.2 + NO = NO.sub.2 + OH                                                                     2.11E12   0.0    -479                                     NO.sub.2 + H = NO + OH                                                                            3.5E14    0.0    1500                                     NO.sub.2 + O = NO + O.sub.2                                                                       1.0E13    0.0    600                                      NO.sub.2 + M = NO + O + M                                                                         1.1E16    0.0    66000                                    NCO + H = NH + CO   5.0E13    0.0    0                                        NCO + O = NO + CO   2.0E13    0.0    0                                        NCO + N = N2 + CO   2.0E13    0.0    0                                        NCO + OH = NO + CO + H                                                                            1.0E13    0.0    0                                        NCO + M = N + CO + M                                                                              3.1E16    -0.5   48000                                    NCO + NO = N.sub.2 O + CO                                                                         1.0E13    0.0    -390                                     NCO + H.sub.2 = HNCO + H                                                                          8.58E12   0.0    9000                                     HNCO + H = NH.sub.2 + CO                                                                          2.0E13    0.0    3000                                     NH + O.sub.2 = HNO + O                                                                            1.0E13    0.0    12000                                    NH + O.sub.2 = NO + OH                                                                            7.6E10    0.0    1530                                     NH + NO = N.sub.2 O + H                                                                           2.4E15    -0.8   0                                        N.sub.2 O + OH = N.sub.2 + HO.sub.2                                                               2.0E12    0.0    10000                                    N.sub.2 O + H = N.sub.2 + OH                                                                      7.6E13    0.0    15200                                    N.sub.2 O + M = N.sub.2 + 0 + M                                                                   1.6E14    0.0    51600                                    N.sub.2 O + 0 = N.sub.2 + O.sub.2                                                                 1.0E14    0.0    28200                                    N.sub.2 O + 0 = NO + NO                                                                           1.0E14    0.0    28200                                    NH + OH = HNO + H   2.0E13    0.0    0                                        NH + OH = N + H.sub.2 O                                                                           5.0E11    0.5    2000                                     NH + N = N.sub.2 + H                                                                              3.0E13    0.0    0                                        NH + H = N + H.sub.2                                                                              1.0E14    0.0    0                                        NH.sub.2 + O = HNO + H                                                                            6.63E14   -0.5   0                                        NH.sub.2 + 0 = NH + OH                                                                            6.75E12   0.0    0                                        NH.sub.2 + OH = NH + H.sub.2 O                                                                    4.0E6     2.0    1000                                     NH.sub.2 + H = NH + H.sub.2                                                                       6.92E13   0.0    3650                                     NH.sub.2 + NO = NNH + OH                                                                          6.40E15   -1.25  0                                        NH.sub.2 + NO = N.sub.2 + H.sub.2 O                                                               6.20E15   -1.25  0                                        NH.sub.3 + OH = NH.sub.2 + H.sub.2 O                                                              2.04E6    2.04   566                                      NH.sub.3 + H = NH.sub.2 + H.sub.2                                                                 6.36E5    2.39   10171                                    NH.sub.3 + O = NH.sub.2 + OH                                                                      2.1E13    0.0    9000                                     NNH = N.sub.2 + H   1.0E4     0.0    0                                        NNH + NO = N.sub.2 + HNO                                                                          5.0E13    0.0    0                                        NNH + H = N.sub.2 + H.sub.2                                                                       1.0E14    0.0    0                                        NNH + OH = N.sub.2 + H.sub.2 O                                                                    5.0E13    0.0    0                                        NNH + NH.sub.2 = N.sub.2 + NH.sub.3                                                               5.0E13    0.0    0                                        NNH + NH = N.sub.2 + NH.sub.2                                                                     5.0E13    0.0    0                                        NNH + O = N.sub.2 O + H                                                                           1.0E14    0.0    0                                        HNO + M = H + NO + M                                                                              1.5E16    0.0    48680                                    H.sub.2 O/10/O2/2/N.sub.2 /2/H.sub.2 /2/                                      HNO + OH = NO + H.sub.2 O                                                                         3.6E13    0.0    0                                        HNO + H = H.sub.2 + NO                                                                            5.0E12    0.0    0                                        HNO + NH.sub.2 = NH.sub.3 + NO                                                                    2.0E13    0.0    1000                                     N + NO = N.sub.2 + O                                                                              3.27E12   0.3    0                                        N + O.sub.2 = NO + O                                                                              6.4E9     1.0    6280                                     N + OH = NO + H     3.8E13    0.0    0                                        NH + O = NO + H     0.200E + 14                                                                             0.00   0                                        2HNO = N.sub.2 O + H.sub.2 O                                                                      0.395E + 13                                                                             0.00   5000                                     HNO + NO = N.sub.2 O + OH                                                                         0.200E + 13                                                                             0.00   26000                                    NH.sub.2 + NH = N.sub.2 H.sub.2 + H                                                               0.500E + 14                                                                             0.00   0                                        2NH = N.sub.2 + 2H  0.254E + 14                                                                             0.00   0                                        NH.sub.2 + N = N.sub.2 + 2H                                                                       0.720E + 14                                                                             0.00   0                                        N.sub.2 H.sub.2 + M = NNH + H + M                                                                 0.500E + 17                                                                             0.00   50000                                    H.sub.2 O/15.0/O.sub.2 /2.0/N.sub.2 /2.0/H.sub.2 /2.0/                        N.sub.2 H.sub.2 + H = NNH + H.sub.2                                                               0.500E + 14                                                                             0.00   1000                                     N.sub.2 H.sub.2 + O = NH.sub.2 + NO                                                               0.100E + 14                                                                             0.00   0                                        N.sub.2 H.sub.2 + O = NNH + OH                                                                    0.200E + 14                                                                             0.00   1000                                     N.sub.2 H.sub.2 + OH = NNH + H.sub.2 O                                                            0.100E + 14                                                                             0.00   1000                                     N.sub.2 H.sub.2 + NO = N.sub.2 O + NH.sub.2                                                       0.300E + 13                                                                             0.00   0                                        N.sub.2 H.sub.2 + NH = NNH + NH.sub.2                                                             0.100E + 14                                                                             0.00   1000                                     N.sub.2 H.sub.2 + NH.sub.2 = NH.sub.3 + NNH                                                       0.100E + 14                                                                             0.00   1000                                     2NH.sub.2 = N.sub.2 H.sub.2 + H.sub.2                                                             0.500E + 12                                                                             0.00   0                                        NH.sub.2 + O.sub.2 = HNO + OH                                                                     0.450E + 13                                                                             0.00   25000                                    O + O.sub.2 + M = O.sub.3 + M                                                                     0.782E + 21                                                                             -2.70  0                                        O + O.sub.3 = O.sub.2 + O.sub.2                                                                   0.282E + 13                                                                             0.00   4094                                     NO + OH + M = HNO.sub.2 + M                                                                       0.233E + 24                                                                             -2.40  0                                        OH + HNO.sub.2 = H.sub.2 O + NO.sub.2                                                             0.108E + 14                                                                             0.00   775                                      NO + O.sub.3 = NO.sub.2 + O.sub.2                                                                 0.121E + 13                                                                             0.00   2782                                     SO.sub.2 + O + M = SO.sub.3 + M                                                                   0.145E + 17                                                                             0.00   2000                                     SO.sub.2 + OH + M = HSO.sub.3 + M                                                                 0.566E + 24                                                                             -2.60  0                                        HSO.sub.3 + O.sub.2 = HO.sub.2 + SO.sub.3                                                         0.783E + 12                                                                             0.00   656                                      CH.sub.3 OH = CH.sub.3 + OH                                                                       0.190E + 17                                                                             0.00   91780                                    CH.sub.3 OH = CH.sub.2 OH + H                                                                     0.154E + 17                                                                             0.00   96790                                    CH.sub.3 OH + OH = CH.sub.2 OH + H.sub.2 O                                                        0.177E + 05                                                                             2.65   -883                                     CH.sub.3 OH + OH = CH.sub.3 O + H.sub.2 O                                                         0.177E + 05                                                                             2.65   -883                                     CH.sub.3 OH + H = CH.sub.2 OH + H.sub.2                                                           0.320E + 14                                                                             0.00   6095                                     CH.sub.3 OH + H = CH.sub.3 O + H.sub.2                                                            0.800E + 13                                                                             0.00   6095                                     CH.sub.3 OH + O = CH.sub.2 OH + OH                                                                0.388E + 06                                                                             2.50   3080                                     CH.sub.3 OH + HO.sub.2 = CH.sub.2 OH + H.sub.2 O.sub.2                                            0.398E + 14                                                                             0.00   19400                                    CH.sub.3 OH + CH.sub.3 = CH.sub.2 OH + CH.sub.4                                                   0.319E + 02                                                                             3.17   7172                                     CH.sub.3 OH + CH.sub.3 = CH.sub.3 O + CH.sub.4                                                    0.145E + 02                                                                             3.10   6935                                     CH.sub.2 OH + CH.sub.2 OH = CH.sub.3 OH + CH.sub.2 O                                              0.120E + 14                                                                             0.00   0                                        CH.sub.2 OH + HCO = CH.sub.3 OH + CO                                                              0.120E + 15                                                                             0.00   0                                        H.sub.2 O.sub.2 (wall) => H.sub.2 O + 0.5 O.sub.2                                                 0.550E + 01                                                                             0.00   2500                                     ______________________________________                                    

Example 1

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        NO                  100 ppm                                                   H.sub.2 O.sub.2     200 ppm                                                   O.sub.2             4.2%                                                      H.sub.2 O           5.4%                                                      N.sub.2             Balance                                                   ______________________________________                                    

The reaction time was 1.6 seconds (s) and the pressure was held at 1atmosphere (atm). The temperature was set at a constant 750 K (477° C.).It was found under these conditions that at the end of 1.6 seconds, theNO was reduced from 100 ppm to 4 ppm. This example illustrates that attemperatures in excess of 650 K (377° C.), oxidation of NO to NO₂ takesplace in the presence of H₂ O₂.

Example 2

In this example, the experiment described in Example 1 was repeated,however, the temperature was 1100 K (827° C.) and the reaction time was1.0 s. It was found under these conditions that at the end of 1.0 s, theNO was reduced from 100 ppm to 80 ppm. This example illustrates that attemperatures of about 1100 K (827° C.) and higher there is no effectiveNO-to-NO₂ conversion.

Example 3

In this example, the experiment described in Example 1 was repeated,however, the temperature was 600 K (327° C.) and the reaction time was2.0 s. It was found under these conditions that at the end of 2.0 s, theNO was reduced from 100 ppm to 70 ppm. This example illustrates that attemperatures of about 600 K (327° C.) and lower there is no effectiveNO-to-NO₂ conversion.

Example 4

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        NO                  200 ppm                                                   H.sub.2 O.sub.2     300 ppm                                                   CO                  24 ppm                                                    O.sub.2             3%                                                        H.sub.2 O           15%                                                       CO.sub.2            8%                                                        N.sub.2             Balance                                                   ______________________________________                                    

The reaction time was 1.0 s and the pressure was held at 1 atm. Thetemperature was set at a constant of 800 K (527° C.). It was found underthese conditions that at the end of 1.0 s., the NO was reduced from 200ppm to 20 ppm. There were no visible CO concentration changes. Thisexample illustrates that at a temperature of about 800 K (527° C.),oxidation of NO to NO₂ takes place in the presence of H₂ O₂, but thetemperature is not high enough for CO oxidation.

Example 5

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        NO                  70 ppm                                                    H.sub.2 O.sub.2     53 ppm                                                    CH.sub.3 OH         52 ppm                                                    CO                  35 ppm                                                    O.sub.2             3%                                                        H.sub.2 O           15%                                                       CO.sub.2            8%                                                        N.sub.2             Balance                                                   ______________________________________                                    

The reaction time was 1.0 s and the pressure was held at 1 atm. Thetemperature was set at a constant 800 K (527° C.). It was found underthese conditions that at the end of 1.0 s, the NO was reduced from 70ppm to 8 ppm. At the same time, the CO concentration increased from 35to 85 ppm. This example illustrates that at a temperature of about 800 K(527° C.), oxidation of NO to NO₂ takes place in the presence of amixture of H₂ O₂ and methanol, but the methanol is converted to CO.

Example 6

In this example, the experiment described in Example 5 was repeated,however, the initial H₂ O₂ concentration was 84 ppm and the initial CH₃OH concentration was 21 ppm with the same H₂ O₂ +CH₃ OH level of 105ppm. It was found under these conditions that the NO was reduced from 70ppm to 17 ppm. At the same time, the CO concentration increased from 35to 58 ppm. This example illustrates that at a temperature of about 800 K(527° C.), oxidation of NO to NO₂ takes place at various H₂ O₂ to CH₃ OHratios, but at any initial CH₃ OH concentration almost all CH₃ OH isconverted to CO.

Example 7

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        SO.sub.3            100 ppm                                                   H.sub.2 O.sub.2     200 ppm                                                   O.sub.2             4.2%                                                      H.sub.2 O           5.4%                                                      N.sub.2             Balance                                                   ______________________________________                                    

The reaction time was 1.3 s and the pressure was held at 1 atm. Thetemperature was set at a constant 840 K (567° C.). It was found underthese conditions that at the end of 1.3 s, about 90 ppm SO₂ was formedfrom the SO₃. This example illustrates that at temperatures of about 840K (567° C.) and higher, SO₃ to SO₂ conversion takes place in thepresence of H₂ O₂.

Example 8

In this example, the experiment described in Example 7 was repeated,however, the temperature was 1100 K (827° C.) and the reaction time was1.0 s. It was found under these conditions that at the end of 1.0 s, theSO₂ was not formed. This example illustrates that at temperatures ofabout 1100 K (827° C.) and higher, there is no effective SO₃ -to-SO₂conversion due to H₂ O₂ injection.

Example 9

In this example, the experiment described in Example 7 was repeated,however, the temperature was 700 K (427° C.) and the reaction time was1.6 s. It was found under these conditions that at the end of 1.6 s, theSO₂ was not formed. This example illustrates that at temperatures ofabout 700 K (427° C.) and lower, there is no effective SO₃ to SO₂conversion.

Example 10

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        CO                  90 ppm                                                    H.sub.2 O.sub.2     180 ppm                                                   O.sub.2             4.2%                                                      H.sub.2 O           5.4%                                                      N.sub.2             Balance                                                   ______________________________________                                    

The reaction time was 1.3 s and the pressure was held at 1 atm. Thetemperature was set at a constant 900 K (627° C.). It was found underthese conditions that at the end of 1.3 s, the CO concentration wasreduced to about 68 ppm. This example illustrates that at a temperatureof about 900 K (627° C.), the CO concentration is reduced in thepresence of H₂ O₂. Under the same conditions but without hydrogenperoxide, the temperature was not high enough for CO oxidation bymolecular oxygen.

Example 11

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        CH.sub.4            90 ppm                                                    H.sub.2 O.sub.2     180 ppm                                                   O.sub.2             4.2%                                                      H.sub.2 O           5.4%                                                      N.sub.2             Balance                                                   ______________________________________                                    

The reaction time was 1.3 s and the pressure was held at 1 atm. Thetemperature was set at a constant of 1000 K (727° C.). It was foundunder these conditions that at the end of 1.3 s, the CH₄ concentrationwas reduced to about 4 ppm. This example illustrates that at atemperature of about 1000 K (727° C.), the CH₄ concentration is reducedin the presence of H₂ O₂. Under the same conditions but without hydrogenperoxide, the temperature was not high enough for CH₄ oxidation bymolecular oxygen.

Example 12

Hydrogen peroxide was injected into a gas stream having the followingcomposition:

    ______________________________________                                        Component           Concentration                                             ______________________________________                                        Hg                  0.1 ppm                                                   SO.sub.3            20 ppm                                                    SO.sub.2            2000 ppm                                                  NO                  200 ppm                                                   H.sub.2 O.sub.2     400 ppm                                                   O.sub.2             3%                                                        CO.sub.2            8%                                                        H.sub.2 O           15%                                                       N.sub.2             Balance                                                   ______________________________________                                    

Reaction (6), Hg+HO₂ →HgO+OH, with a rate constant of k=10¹³.5 cc/mol.s,was added to the mechanism set forth in Table I at temperatures of 800 K(527° C.) and 900 K (627° C.). A substantially complete Hg removal wasfound within a reaction time of about 0.04 and 0.01 s, respectively. Atthe same time, the NO-to-NO₂ and SO₃ -to-SO₂ conversions took place.This example illustrates that Hg removal is effective at temperaturesbetween about 700 K (427° C.) and 1000 K (727° C.).

Example 13

In this example, the model experiment described in Example 12 was run atdifferent temperatures and at varying rate constants for the reaction(6). It was found that the variation of the rate constant within anorder of magnitude does not significantly affect Hg removal. While theefficiency of Hg removal decreases at lower rate constants, even at arate constant of 10¹¹.5 cc/mol.s (which is two orders of magnitude lowerthan the value used in Example 12), substantially complete Hg removalcan be achieved at a temperature of 800 K (527° C.).

The invention may be embodied in other specific forms without departingfrom its spirit or essential characteristics. The described embodimentsare to be considered in all respects only as illustrative and notrestrictive. The scope of the invention is, therefore, indicated by theappended claims rather than by the foregoing description. All changesthat come within the meaning and range of equivalency of the claims areto be embraced within their scope.

What is claimed and desired to be secured by United States LettersPatent is:
 1. A method for removing gas components from a combustionflue gas including one or more of NO, SO₃, CO, light hydrocarbons, andmercury vapor, the method comprising the step of contacting thecombustion flue gas with an injection liquid including atomized dropletsof a mixture of aqueous hydrogen peroxide solution having aconcentration from about 1% to about 50% and methanol in an amount suchthat the mole ratio of the sum of the hydrogen peroxide and methanol tothe sum of any such NO, SO₃, CO, light hydrocarbons and mercury vaporcontained in the combustion flue gas is in a range from about 0.5 toabout 2, and the concentration of the mixture of hydrogen peroxide andmethanol in the combustion flue gas is less than about 1000 ppm, at aflue gas temperature in a range from about 650 K (377° C.) to about 1100K (827° C.), wherein the mixture of hydrogen peroxide and methanolsubstantially converts any NO, SO₃, CO, light hydrocarbons, and mercuryvapor in the combustion flue gas to NO₂, SO₂, CO₂, and HgO by a chainpropagating reaction in the combustion flue gas.
 2. The method of claim1, wherein the combustion flue gas further comprises initialconcentrations of carbon dioxide, water and oxygen.
 3. The method ofclaim 1, wherein the mole ratio of the mixture of hydrogen peroxide andmethanol to NO is in a range from about 0.5 to about
 2. 4. The method ofclaim 1, wherein the initial concentration of SO₃ in the flue gas is upto about 100 ppm.
 5. The method of claim 1, wherein the initialconcentration of CO in the flue gas is up to about 500 ppm.
 6. Themethod of claim 1, wherein the initial concentration of lighthydrocarbons in the flue gas is up to about 1000 ppm.
 7. The method ofclaim 1, wherein the initial concentration of Hg in the flue gas is lessthan about 1 ppm.
 8. The method of claim 1, wherein the mixture ofhydrogen peroxide and methanol is injected into the flue gas upstream ofa particulate control device.
 9. The method of claim 1, wherein thereaction time of the mixture of hydrogen peroxide and methanol and thegas components is between about 0.01 to about 5 seconds.
 10. The methodof claim 1, further comprising the step of removing the NO₂, SO₂, andHgO from the combustion flue gas.
 11. The method of claim 1, wherein themixture of hydrogen peroxide and methanol is propelled into the flue gasby a jet of gas.
 12. The method of claim 1, wherein the flue gastemperature is in the range from about 700 K (427° C.) to about 1000 K(727° C.).
 13. A method for removing gas components from a combustionflue gas including one or more of NO, SO₃, CO, light hydrocarbons, andmercury vapor, the method comprising the step of contacting thecombustion flue gas with an injection liquid including atomized dropletsof a mixture of aqueous hydrogen peroxide solution having aconcentration from about 10% to about 30% and methanol in an amount suchthat the mole ratio of the sum of the hydrogen peroxide and methanol tothe sum of any such NO, SO₃, CO, light hydrocarbons and mercury vaporcontained in the combustion flue gas is in a range from about 0.5 toabout 2, and the concentration of the mixture of hydrogen peroxide andmethanol in the combustion flue gas is less than about 500 ppm, at aflue gas temperature in a range from about 650 K (377° C.) to about 1100K (827° C.), wherein the mixture of hydrogen peroxide and methanolsubstantially converts any NO, SO₃, CO, light hydrocarbons, and mercuryvapor in the combustion flue gas to NO₂, SO₂, CO₂, and HgO by a chainpropagating reaction in the combustion flue gas.
 14. The method of claim13, wherein the combustion flue gas further comprises initialconcentrations of carbon dioxide, water, and oxygen.
 15. The method ofclaim 13, wherein the mole ratio of the mixture of hydrogen peroxide andmethanol to NO is in a range from about 0.5 to about
 2. 16. The methodof claim 13, wherein the initial concentration of SO₃ in the flue gas isup to about 100 ppm.
 17. The method of claim 13, wherein the initialconcentration of CO in the flue gas is up to about 500 ppm.
 18. Themethod of claim 13, wherein the initial concentration of lighthydrocarbons in the flue gas is up to about 1000 ppm.
 19. The method ofclaim 13, wherein the initial concentration of mercury in the flue gasis less than about 1 ppm.
 20. The method of claim 13, wherein themixture of hydrogen peroxide and methanol is injected into the flue gasupstream of a particulate control device.
 21. The method of claim 13,wherein the reaction time of the mixture of hydrogen peroxide andmethanol and the gas components is between about 0.01 to about 5seconds.
 22. The method of claim 13, further comprising the step ofremoving the NO₂, SO₂, and HgO from the combustion flue gas.
 23. Themethod of claim 13, wherein the mixture of hydrogen peroxide andmethanol is atomized and propelled into the flue gas by a jet of gas.24. The method of claim 13, wherein the flue gas temperature is in therange from about 700 K (427° C.) to about 1000 K (727° C.).